What is the equation of the tangent line to the function
f(x) = x2 - 2x + 1 at a = 0?

Question

What is the equation of the tangent line to the function 
f(x) = x2 - 2x + 1 at a = 0?

anonymous · Answer

and a brief explanation please.

amistre64 · Answer

at a=0?

anonymous · Answer

The formula is:
$$y-y _{0}=f \prime(x)(x-x _{0})$$The formula is analogous to the formula of a line in point-slope form. The slope of the tangent line to a curve has two properties. It passes through the point of interest (in your case (0,1)) and it has the same rate of change (ie., slope) as the curve at that point.  So we have,$$y-1=(2(0)-2)(x-0)$$This becomes,$$y=-2x+1$$

anonymous · Answer

Should say "the tangent line to a curve has two properties..."

anonymous · Answer

to get the equation of the tangent line, we need to things, a point on the line, and the slope of the line.
For the slope, we can take the derivative of the function and plug in x= 0 (since you want the equation of the tangent line at x = 0):
$$f(x) = x^2-2x+1 \Rightarrow f'(x) = 2x-2 \Rightarrow f'(0) = 2(0)-2 = -2$$
So the slope is -2.
Now for the point, lets plug x = 0 into the original function (not the derivative) to see what the value of y is:
$$f(x) = x^2-2x+1 \Rightarrow f(0) = 0^2-0*2+1 = 1$$
So the point is (0,1)
Now that you have a slope and point, use the formula:
$$y-y_1=m(x-x_1)$$
to get the equation of the line:
$$y-1 = -2(x-0) \Rightarrow y = -2x+1$$

amistre64 · Answer

or; y = f'(a)x -f'(a)a + f(a) .. maybe